如何编写不使用typedefs即可返回函数指针的函数指针？

Question

I saw this , but it never specified how to declare a function pointer that returns a pointer-to-a-function (it simply defined how to create a function returning a function pointer).

While it would probably be wiser to use typedefs, I am interested in the syntax to accomplish this without typedefs.

This is the closest I got:

struct function_hash_table {
  unsigned int(*(*getFunction)(Type, Type, Type))( char *name );
}

(What type actually is is rather irrelevant to this question).

But, when I try to call it like this:

hash_table.getFunction( "Test" );

I get errors for too few parameters and wrong parameter types.

To Clarify:

unsigned int(*hash_get_function(Type, Type, Type))(char *name)

Works good for actual functions and explained in the link. However, how am I suppose to declare a function pointer that refers to such a definition?

Answer 1

how to declare a function pointer that returns a pointer-to-a-function

Without using typedef you can declare as

Type (* (*funcPtr)(Type, Type) ) (Type); 

Declaration read as: funcPtr is a pointer to a function that expects two arguments of type Type and returns a pointer to a function that expects an argument of type Type and returns Type .

If you wan to declare a function that return a pointer then

Type (* func(Type, Type) ) (Type);   

Declaration read as: func is a function that accepts two parameters of type Type and returns a pointer to function that accepts an argument of type Type and returns Type .

Now you can assign func to funcPtr

functPtr = func;